Via reddit, a challenge to solve:

$y = y’ + y’’ + y’’’ + \dots$

Once you’ve stopped running away and hiding, I’ll show you the solution they suggested.

The trick is to notice that $y’ = y’’ + y’’’ + y^{(4)} + \dots$, so $y = y’ + (y’)$, which is straightforward to solve:

  • $y = 2y’$
  • $y = Ae^{x/2}$.

It’s a good idea to check that it works:

  • $y’ = \frac{1}{2}y$
  • $y’’ = \frac{1}{4}y$
  • $\dots$

The right-hand side is then $y\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8}+\dots\right)$, and the bracket evaluates to 1. Boom.

But wait a moment

We’ve found a solution. How do we know it’s the only one?

What if you notice that $y’’ = y’’’ + y^{(4)} + \dots$ and get to $y = y’ + 2y’’$?

That solves as $y = Ae^{x/2} + Be^{-x}$. Isn’t that a solution?

Actually, not if $B$ is non-zero – the sum of the derivatives doesn’t converge – but all the same, you could (in principle) do this with any of the derivatives and is there any guarantee that there’s no other solution?

Stand back.

I’m going to try something I haven’t done in 25 years ((I mean, obviously I have, I did it just before writing up the post. But the point stands.)): Laplace transforms.

The tl;dr is that you convert every term of your differential equation (in “$t$-space”, even though we’re really using $x$ here) into a different term in $s$-space, do your magic there and convert back.

Do not worry about what $t$- and $s$- space are, or how the transform works; that’s not the purpose of this article.

The Laplace transform of the ODE:

$0 = -y + y’ + y’’ + y’’’ + \dots$


$\begin{aligned}[t]0 = &-Y(s) &
& + sY(s) &- y(0)
& + s^2Y(s) &- sy(0) &- y’(0)
& + s^3Y(s) &- s^2y(0) & -sy’(0) & -y’‘(0)
& + \dots \end{aligned}$

At first glance, that looks awful. And then you spot that the columns (apart, slightly, from the first) are geometric sequences! This all reduces to

$0=\left(-2 + \frac{s}{1-s}\right)Y(s) - \frac{1}{1-s}\left(y(0) + y’(0) + \dots\right)$

(As long as $s$ is small enough, I guess.)

Now, we don’t really care about the $y(0)$s and so on, and we can just treat them as a single constant. We can also simplify the bracket on the first term:

$0 = \frac{2s-1}{1-s}Y(s) - \frac{1}{1-s}A$

And as long as $s$ isn’t $1$ ((I don’t even know if that makes sense, honestly, like I said, it’s been 25 years)), the solution to that is $Y(s) = \frac{A}{2s-1}$.

The inverse Laplace transform of that is $y(x) = Ae^{x/2}$, as we had before.

I’m still a bit uneasy about the convergence issues on $s$, but I’m going to pretend they don’t matter. At any rate, I’ve convinced myself that this is the only solution, and that’s the main thing.